已知a、b、c是非零实数,且a^2+b^2+c^2=1,a(1⼀b+1⼀c)+b(1⼀a+1⼀c)+c(1⼀a+1⼀b)=-3,求a+b+c的值

2025-06-28 04:07:39
推荐回答(1个)
回答1:

a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)=-3
a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)+a*1/a+b*1/b+c*1/c=0
a(1/a+1/b+1/c)+b(1/b+1/a+1/c)+c(1/c+1/a+1/b)=0
(a+b+c)(1/a+1/b+1/c)=0
将1置换成a^2+b^2+c^2得:
(a+b+c)*(a+b+c+2(a^2+b^2+c^2))=0
(a+b+c)*(a+b+c+2)=0
将a+b+c看做一个整体
a+b+c=0或-2